How to calculate the temperature from heat of neutralisation?

The heat of neutralization of $$\ce{HCl(aq)}$$ by $$\ce{NaOH}$$ is $$\pu{-55.9 kJ/mol}$$ $$\ce{H2O}$$ produced. If $$\pu{50ml}$$ of $$\pu{1.6M}$$ $$\ce{NaOH}$$ at $$\pu{25.15^\circ C}$$ is added to $$\pu{25ml}$$ of $$\pu{1.79M}$$ $$\ce{HCl}$$ at $$\pu{26.34^\circ C}$$ in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?

I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)

I need a detailed explanation to understand! I'm not sure which one is the initial temperature? $$\pu{25.15^\circ C}$$ of $$\ce{NaOH}$$ or $$\pu{26.34^\circ C}~\ce{HCl}$$?

First, you need to recognize what the limiting reagent is. Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($$\pu{25mL}$$ of $$\pu{1.79M}$$ is fewer moles than $$\pu{50 mL}$$ of $$\pu{1.6M}$$), HCl is the limiting reagent.
Second, given the amount of limiting reagent, how much heat will be released (or absorbed). $$\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$$
Third, you need to approximate that the solution has the heat capacity of water, which is $$\pu{4.18 kJ/K L}$$.
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$
So finally calculate the amount the temperature of $$\pu{75 ml}$$ of "water" will increase when $$\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$$ of heat is added, and add this to $$\pu{25.55 ^\circ C}$$.
• If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. As long as there is no phase changes. – LDC3 Nov 18 '14 at 2:20